Question

in the following triangle, find length bw. round to the nearest tenth.
bw =

Explanation:

Step1: Identify the geometric theorem

This is a right triangle with an altitude to the hypotenuse, so we can use the geometric mean theorem (or altitude-on-hypotenuse theorem), which states that in a right triangle, the length of a leg is the geometric mean of the length of the hypotenuse segment adjacent to that leg and the length of the hypotenuse. Let \( BW = x \), \( WS = 16.1 \), and \( BN = 12 \). The hypotenuse \( BS=x + 16.1\). By the geometric mean theorem, \( BN^{2}=BW\times BS \), so \( 12^{2}=x(x + 16.1) \).

Step2: Solve the quadratic equation

We have the equation \( 144=x^{2}+16.1x \), which can be rewritten as \( x^{2}+16.1x - 144 = 0 \). Using the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 1 \), \( b = 16.1 \), and \( c=- 144 \). First, calculate the discriminant \( \Delta=b^{2}-4ac=(16.1)^{2}-4\times1\times(-144)=259.21 + 576 = 835.21 \). Then \( \sqrt{\Delta}=\sqrt{835.21}=28.9 \). So \( x=\frac{-16.1\pm28.9}{2} \). We take the positive root because length can't be negative: \( x=\frac{-16.1 + 28.9}{2}=\frac{12.8}{2}=6.4 \)? Wait, no, maybe I made a mistake. Wait, actually, the geometric mean theorem can also be applied as \( BW=\frac{BN^{2}}{BS} \)? Wait, no, let's re - examine the triangle. Let's denote the right triangle \( \triangle BNS \) with right angle at \( N \), and \( NW\perp BS \). Then \( \triangle BNW\sim\triangle BNS \sim\triangle NWS \). So \( \frac{BW}{BN}=\frac{BN}{BS} \), so \( BW=\frac{BN^{2}}{BS} \). Wait, but we need to find \( BS \)? Wait, no, maybe I misread the diagram. Wait, the length of \( WS = 16.1 \), and we need to find \( BW \). Wait, actually, another way: in a right triangle, if we let the hypotenuse be \( BS \), and the leg \( BN = 12 \), and the segment \( WS=16.1 \), and \( BW=x \), then by the geometric mean theorem, \( BN^{2}=BW\times BS \), and \( BS=BW + WS=x + 16.1 \). So \( 12^{2}=x(x + 16.1) \), \( x^{2}+16.1x-144 = 0 \). But when we calculate the discriminant: \( 16.1^{2}=259.21 \), \( 4\times1\times144 = 576 \), so discriminant is \( 259.21+576 = 835.21 \), square root of 835.21 is 28.9. Then \( x=\frac{-16.1\pm28.9}{2} \). Taking the positive root: \( x=\frac{-16.1 + 28.9}{2}=\frac{12.8}{2}=6.4 \)? Wait, that seems low. Wait, maybe the length of \( WS \) is 16.1, and \( BS=BW + WS \), but maybe I got the segments wrong. Wait, maybe the hypotenuse is \( BS \), and \( BN = 12 \), \( NS \) is another leg, and \( NW \) is the altitude. Wait, alternatively, maybe the triangle is such that \( \triangle BNW \) and \( \triangle BNS \) are similar, so \( \frac{BW}{BN}=\frac{BN}{BS} \), but if we consider that \( BS=BW + WS \), and \( WS = 16.1 \), but maybe I made a mistake in the theorem application. Wait, let's use another approach. Let's assume that \( BS \) is the hypotenuse, and we know that in a right triangle, the altitude to the hypotenuse \( NW \) satisfies \( NW^{2}=BW\times WS \), but we don't know \( NW \). Wait, maybe the length of \( BN = 12 \), and we can find \( BS \) first using Pythagoras? Wait, no, we don't know \( NS \). Wait, maybe the diagram is such that \( BS \) is the hypotenuse, \( BN = 12 \), and \( WS=16.1 \), and \( BW \) is the segment we need to find. Wait, perhaps the correct formula is \( BW=\frac{BN^{2}}{BS} \), but we need to find \( BS \). Wait, maybe I misread the problem. Wait, the problem says "find length \( BW \)", and the diagram has \( BN = 12 \), \( WS = 16.1 \), and right angles at \( N \) and \( W \). So \( \triangle BNW \) and \( \triangle BNS \) are similar. So \( \frac{BW}{BN}=\frac{BN}{BS} \), so \( BW=\frac{BN^{2}}{BS} \), and \( BS=BW + WS \). Let \( BW=x \), then \( x=\frac{12^{2}}{x + 16.1} \), \( x(x + 16.1)=144 \), \( x^{2}+16.1x - 144 = 0 \). Solving this quadratic equation: \( x=\frac{-16.1\pm\sqrt{16.1^{2}+4\times144}}{2}=\frac{-16.1\pm\sqrt{259.21 + 576}}{2}=\frac{-16.1\pm\sqrt{835.21}}{2}=\frac{-16.1\pm28.9}{2} \). Taking the positive root: \( x=\frac{-16.1 + 28.9}{2}=\frac{12.8}{2}=6.4 \)? Wait, that seems correct. Wait, but let's check with another method. If we consider that in a right triangle, the length of the leg is the geometric mean of the hypotenuse and the adjacent segment. So if \( BN \) is a leg, \( BW \) is the adjacent segment to \( BN \) on the hypotenuse \( BS \), then \( BN^{2}=BW\times BS \). Let \( BW=x \), \( BS=x + 16.1 \), so \( 12^{2}=x(x + 16.1) \), \( x^{2}+16.1x - 144 = 0 \). Using quadratic formula: \( x=\frac{-16.1+\sqrt{16.1^{2}+4\times144}}{2}=\frac{-16.1 + 28.9}{2}=6.4 \). So \( BW = 6.4 \)? Wait, but let's verify. If \( BW = 6.4 \), then \( BS=6.4 + 16.1 = 22.5 \), and \( 12^{2}=144 \), \( 6.4\times22.5 = 144 \), which is correct. So the length of \( BW \) is 6.4.

Answer:

\( 6.4 \)